Question: Simplify and expand the following expression: $ \dfrac{3k + 6}{3k + 1}-\dfrac{3k - 7}{4k + 10} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3k + 1)(4k + 10)$ Multiply the first term by $\dfrac{4k + 10}{4k + 10}$ $ \begin{align*} \dfrac{3k + 6}{3k + 1} \times \dfrac{4k + 10}{4k + 10} & = \dfrac{(3k + 6)(4k + 10)}{(3k + 1)(4k + 10)} \\ & = \dfrac{12k^2 + 54k + 60}{(3k + 1)(4k + 10)}\end{align*} $ Multiply the second term by $\dfrac{3k + 1}{3k + 1}$ $ \begin{align*} \dfrac{3k - 7}{4k + 10} \times \dfrac{3k + 1}{3k + 1} & = \dfrac{(3k - 7)(3k + 1)}{(4k + 10)(3k + 1)} \\ & = \dfrac{9k^2 - 18k - 7}{(4k + 10)(3k + 1)}\end{align*} $ Now we have: $ = \dfrac{12k^2 + 54k + 60}{(3k + 1)(4k + 10)} - \dfrac{9k^2 - 18k - 7}{(4k + 10)(3k + 1)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{12k^2 + 54k + 60 - (9k^2 - 18k - 7)}{(3k + 1)(4k + 10)} $ $ = \dfrac{12k^2 + 54k + 60 - 9k^2 + 18k + 7}{(3k + 1)(4k + 10)} $ $ = \dfrac{3k^2 + 72k + 67}{(3k + 1)(4k + 10)}$ Expand the denominator: $ = \dfrac{3k^2 + 72k + 67}{12k^2 + 34k + 10}$